Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k - 5}{-9k^2 + 9k} \div \dfrac{k - 5}{k^2 + 5k - 6} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k - 5}{-9k^2 + 9k} \times \dfrac{k^2 + 5k - 6}{k - 5} $ First factor the quadratic. $n = \dfrac{k - 5}{-9k^2 + 9k} \times \dfrac{(k - 1)(k + 6)}{k - 5} $ Then factor out any other terms. $n = \dfrac{k - 5}{-9k(k - 1)} \times \dfrac{(k - 1)(k + 6)}{k - 5} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (k - 5) \times (k - 1)(k + 6) } { -9k(k - 1) \times (k - 5) } $ $n = \dfrac{ (k - 5)(k - 1)(k + 6)}{ -9k(k - 1)(k - 5)} $ Notice that $(k - 5)$ and $(k - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(k - 5)}(k - 1)(k + 6)}{ -9k\cancel{(k - 1)}(k - 5)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $n = \dfrac{ \cancel{(k - 5)}\cancel{(k - 1)}(k + 6)}{ -9k\cancel{(k - 1)}\cancel{(k - 5)}} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $n = \dfrac{k + 6}{-9k} $ $n = \dfrac{-(k + 6)}{9k} ; \space k \neq 1 ; \space k \neq 5 $